Notes

Chapter 6 Normal Probability Distributions 6-1 Review and Preview 6-2 The Standard Normal Distribution 6-3 Applications of Normal Distributions

6-4 Sampling Distributions and Estimators 6-5 The Central Limit Theorem

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The normal probability distribution has three properties: 1. It’s graph is bell-shaped.

2. It’s mean is equal to µ 3. It’s standard deviation is equal to σ

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Density Curve A density curve is the graph of a continuous probability distribution. It must satisfy the following properties:

1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve cannot fall below the x-axis.) Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Area and Probability Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.

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Standard Normal Distribution The standard normal distribution is a normal probability distribution with   0 and   1 . The total area under its density curve is equal to 1.

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Finding Probabilities When Given z-scores

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Z-table

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Using Z-table 1. It is designed only for the standard normal distribution, which has a mean of 0 and a standard deviation of 1. 2. It is on two pages, with one page for negative z-scores and the other page for positive z-scores. 3. Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score.

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Using Z-Table 4. When working with a graph, avoid confusion between z-scores and areas. z Score Distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Z- table

Area Region under the curve; refer to the values in the body of Z-table. 5. The part of the z-score denoting hundredths is found across the top. Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Example 1 P( z  1.27) 

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Look at Z-Table

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Example 1 - cont P( z  1.27)  0.8980

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Example - 2 P( z  1.23)  0.8907

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Example 3 z between –2.00 and 1.50 degrees, find the area.

P( z  2.00)  0.0228 P( z  1.50)  0.9332 P(2.00  z  1.50)  0.9332  0.0228  0.9104

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Notation P ( a  z  b) denotes the probability that the z score is between a and b.

P( z  a) denotes the probability that the z score is greater than a.

P( z  a) denotes the probability that the z score is less than a.

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Finding a z Score When Given a Probability Using a Z-Table 1. Draw a bell-shaped curve and identify the region under the curve that corresponds to the given probability. If that region is not a cumulative region from the left, work instead with a known region that is a cumulative region from the left. 2. Using the cumulative area from the left, locate the closest probability in the body of Z-Table and identify the corresponding z score.

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Finding z Scores When Given Probabilities 5% or 0.05

(z score will be positive)

Finding the 95th Percentile Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Finding z Scores When Given Probabilities - cont 5% or 0.05

(z score will be positive)

1.645

Finding the 95th Percentile Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Finding z Scores When Given Probabilities - cont

(One z score will be negative and the other positive)

Finding the Bottom 2.5% and Upper 2.5% Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Finding z Scores When Given Probabilities - cont

(One z score will be negative and the other positive)

Finding the Bottom 2.5% and Upper 2.5% Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Recap In this section we have discussed:

 Density curves.  Relationship between area and probability.  Standard normal distribution.  Using Table z- table

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6.3 Application of Normal Distributions

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This section presents methods for working with normal distributions that are not standard. That is, the mean is not 0 or the standard deviation is not 1, or both. The key concept is that we can use a simple conversion that allows us to standardize any normal distribution so that the same methods of the previous section can be used.

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Conversion Formula

z

x



Round z scores to 2 decimal places

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Converting to a Standard Normal Distribution

z

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x



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Example 1 Assuming that the weights of the men in this class are normally distributed with a mean of 172 pounds and standard deviation of 29 pounds. If one man is

randomly selected, what is the probability he weighs less than 174 pounds?

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Example - cont   172   29

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174  172 z  0.07 29

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Example - cont   172   29

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P( x  174lb.)  P ( z  0.07)  0.5279

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Note! 1. Don’t confuse z scores and areas. z scores are distances along the horizontal scale, but areas are regions under the normal curve. Z-Table lists z scores in the left column and across the top row, but areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is located in the left half of the normal distribution. 4. Areas (or probabilities) are positive or zero values, but they are never negative.

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Procedure for Finding Values Using Z-Table and Formula 1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value(s) being sought. 2. Use Z-Table to find the z score corresponding to the cumulative left area bounded by x. Refer to the body of Z- Table to find the closest area, then identify the corresponding z score. 3. Using Formula, enter the values for   and the z score found in step 2, then solve for x.

x    (z  )

(Another form of Formula )

(If z is located to the left of the mean, be sure that it is a negative number.) 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem. Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Example – Lightest and Heaviest Use the data from the previous example to determine what weight separates the lightest 99.5% from the heaviest 0.5%?

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Example – Lightest and Heaviest - cont x    (z  ) x  172  (2.575  29) x  246.675(247 rounded )

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Example – Lightest and Heaviest - cont The weight of 247 pounds separates the lightest 99.5% from the heaviest 0.5%

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Recap In this section we have discussed:

 Non-standard normal distribution.  Converting to a standard normal distribution.  Procedures for finding values using Z-Table and Formula .

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6.4 Sampling Distribution and Estimators

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Definition The sampling distribution of a statistic (such as the sample mean or sample proportion) is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population. (The sampling distribution of a statistic is typically represented as a probability distribution in the format of a table, probability histogram, or formula.)

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Definition The sampling distribution of the mean is the distribution of sample means, with all samples having the same sample size n taken from the same population. (The sampling distribution of the mean is typically represented as a probability distribution in the format of a table, probability histogram, or formula.)

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Properties  Sample means target the value of the population mean. (That is, the mean of the sample means is the population mean. The expected value of the sample mean is equal to the population mean.)

 The distribution of the sample means tends to be a normal distribution.

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Definition The sampling distribution of the variance is the distribution of sample variances, with all samples having the same sample size n taken from the same population. (The sampling distribution of the variance is typically represented as a probability distribution in the format of a table, probability histogram, or formula.)

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Properties  Sample variances target the value of the population variance. (That is, the mean of the sample variances is the population variance. The expected value of the sample variance is equal to the population variance.)

 The distribution of the sample variances tends to be a distribution skewed to the right.

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Definition The sampling distribution of the proportion is the distribution of sample proportions, with all samples having the same sample size n taken from the same population.

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Definition We need to distinguish between a population proportion p and some sample proportion: p = population proportion

pˆ = sample proportion

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Properties  Sample proportions target the value of the population proportion. (That is, the mean of the sample proportions is the population proportion. The expected value of the sample proportion is equal to the population proportion.)  The distribution of the sample proportion tends to be a normal distribution.

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Unbiased Estimators

Sample means, variances and proportions are unbiased estimators. That is they target the population parameter.

These statistics are better in estimating the population parameter. Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Biased Estimators Sample medians, ranges and standard deviations are biased estimators. That is they do NOT target the population parameter.

Note: the bias with the standard deviation is relatively small in large samples so s is often used to estimate. Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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example

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Why Sample with Replacement? Sampling without replacement would have the very practical advantage of avoiding wasteful duplication whenever the same item is selected more than once. However, we are interested in sampling with replacement for these two reasons:

1. When selecting a relatively small sample form a large population, it makes no significant difference whether we sample with replacement or without replacement. 2. Sampling with replacement results in independent events that are unaffected by previous outcomes, and independent events are easier to analyze and result in simpler calculations and formulas. Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Caution

Many methods of statistics require a simple random sample. Some samples, such as voluntary response samples or convenience samples, could easily result in very wrong results.

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Recap In this section we have discussed:

 Sampling distribution of a statistic.  Sampling distribution of the mean.  Sampling distribution of the variance.  Sampling distribution of the proportion.  Estimators.

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6.5 The Central Limit Theorem

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Key Concept The Central Limit Theorem tells us that for a population with any distribution, the distribution of the sample means approaches a normal distribution as the sample size increases. The procedure in this section form the foundation for estimating population parameters and hypothesis testing.

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Central Limit Theorem Given: 1. The random variable x has a distribution (which may or may not be normal) with mean  and standard deviation  . 2. Simple random samples all of size n are selected from the population. (The samples are selected so that all possible samples of the same size n have the same chance of being selected.)

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Central Limit Theorem – cont. Conclusions: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means is the population mean  . 3. The standard deviation of all sample means is  / n .

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Practical Rules Commonly Used 1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets closer to a normal distribution as the sample size n becomes larger. 2. If the original population is normally distributed, then for any sample size n, the sample means will be normally distributed (not just the values of n larger than 30). Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Notation the mean of the sample means

x   the standard deviation of sample mean

x 

 n

(often called the standard error of the mean) Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Important Point As the sample size increases, the sampling distribution of sample means approaches a normal distribution.

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Example – Assume the population of weights of men is normally distributed with a mean of 172 lb and a standard deviation of 29 lb. a) Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb. b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Example – cont a) Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb.

175  172 z  0.10 29

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Example – cont b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb

175  172 z  0.46 29 20

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Example - cont a) Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb.

P( x  175)  0.4602 b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb

P( x  175)  0.3228 It is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate from the mean. Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Correction for a Finite Population When sampling without replacement and the sample size n is greater than 5% of the finite population of size N (that is, n  0.05 N ), adjust the standard deviation of sample means by multiplying it by the finite population correction factor:

x 

 n

N n N 1

finite population correction factor Copyright © 2010, 2007, 2004 Pearson Education, Inc.

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Recap

In this section we have discussed:

 Central limit theorem.  Practical rules.  Effects of sample sizes.  Correction for a finite population.

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